使用ROSALIND平台进行Python实战练习

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Python Village

• INI1--Installing Python

  Problem

  After downloading and installing Python, type import this into the Python command line and see what happens.

• # 安装Python并在命令行模式下输入import this查看Python之禅
  >>> import this
  The Zen of Python, by Tim Peters
  
  Beautiful is better than ugly.
  Explicit is better than implicit.
  Simple is better than complex.
  Complex is better than complicated.
  Flat is better than nested.
  Sparse is better than dense.
  Readability counts.
  Special cases aren't special enough to break the rules.
  Although practicality beats purity.
  Errors should never pass silently.
  Unless explicitly silenced.
  In the face of ambiguity, refuse the temptation to guess.
  There should be one-- and preferably only one --obvious way to do it.
  Although that way may not be obvious at first unless you're Dutch.
  Now is better than never.
  Although never is often better than *right* now.
  If the implementation is hard to explain, it's a bad idea.
  If the implementation is easy to explain, it may be a good idea.
  Namespaces are one honking great idea -- let's do more of those!
  

• INI2--Variables and Some Arithmetic

  Problem

  Given: Two positive integers aa and bb, each less than 1000.

  Return: The integer corresponding to the square of the hypotenuse of the right triangle whose legs have lengths a and b.

• # 求边长为a和b的直角三角形的斜边的平方和
  >>> a = 3
  >>> b = 5
  >>> c = a**2 + b**2
  >>> print(c)
  34
  

• INI3--Strings and Lists

  Problem

  Given: A string s of length at most 200 letters and four integers a, b, c and d.

  Return: The slice of this string from indices a through b and c through
  d (with space in between), inclusively. In other words, we should include elements s[b] and s[d] in our slice.

  # 从一个字符串中按指定索引位置提取子字符串
  >>> s = "WPedostibesprNckwovQydlBbDTX2atKZFiQWkgoppnjtPBJG8VuxVnWH5mDZIClB9Kgf3Wz5U1imodestusKXvI3Zs5gLS4W8r43ZCYQgxGeCv1eWnQveLa9PJZNo3lzjWdmDrfIN7dA1EhOC0NRVuRtHcPaHVu5snowPwjjBUNeDvH0yka33OxLCACg7d0Q."
  >>> a,b,c,d = 1,10,76,83
  >>> print(s[a:b+1])
  Pedostibes
  >>> print(s[c:d+1])
  modestus
  

• INI4--Conditions and Loops

  Problem

  Given: Two positive integers a and b (a < b < 10000).

  Return: The sum of all odd integers from a through b, inclusively.

  # 计算a和b之间所有奇数的累加和
  >> a,b = 4908, 9604
  >>> sum = 0
  >>> for i in range(a,b+1):
  ...     if i % 2 == 1:  # 判断i是否为奇数
  ...             sum += i
  ... 
  >>> print(sum)
  17037088
  

• INI5--Working with Files

  Problem

  Given: A file containing at most 1000 lines.

  Return: A file containing all the even-numbered lines from the original file. Assume 1-based numbering of lines.

  # 读入一个文本文件，循环输出偶数行
  >>> with open("rosalind_ini5.txt","r") as fh:
  ...     for i, line in enumerate(fh):
  ...             if i % 2 == 1: # i是从0开始的
  ...                     print(line,end="")
  ... 
  Some things in life are bad, they can really make you mad
  Other things just make you swear and curse
  When you're chewing on life's gristle, don't grumble give a whistle
  This will help things turn out for the best
  Always look on the bright side of life
  Always look on the right side of life
  If life seems jolly rotten, there's something you've forgotten
  And that's to laugh and smile and dance and sing
  When you're feeling in the dumps, don't be silly, chumps
  Just purse your lips and whistle, that's the thing
  So, always look on the bright side of death
  Just before you draw your terminal breath
  Life's a counterfeit and when you look at it
  Life's a laugh and death's the joke, it's true
  You see, it's all a show, keep them laughing as you go
  Just remember the last laugh is on you
  Always look on the bright side of life
  And always look on the right side of life
  Always look on the bright side of life
  And always look on the right side of life
  

• INI6--Dictionaries

  Problem

  Given: A file containing at most 1000 lines.

  Return: A file containing all the even-numbered lines from the original file. Assume 1-based numbering of lines.

  # 统计一个字符串中每个单词出现的频数，以空格分割
  >>> s = "When I find myself in times of trouble Mother Mary comes to me Speaking words of wisdom let it be And in my hour of darkness she is standing right in front of me Speaking words of wisdom let it be Let it be let it be let it be let it be Whisper words of wisdom let it be And when the broken hearted people living in the world agree There will be an answer let it be For though they may be parted there is still a chance that they will see There will be an answer let it be Let it be let it be let it be let it be There will be an answer let it be Let it be let it be let it be let it be Whisper words of wisdom let it be Let it be let it be let it be let it be Whisper words of wisdom let it be And when the night is cloudy there is still a light that shines on me Shine until tomorrow let it be I wake up to the sound of music Mother Mary comes to me Speaking words of wisdom let it be Let it be let it be let it be yeah let it be There will be an answer let it be Let it be let it be let it be yeah let it be Whisper words of wisdom let it be"
  >>> mydict = {}
  >>> for word in s.split(" "):
  ...     if word in mydict:
  ...             mydict[word] += 1
  ...     else:
  ...             mydict[word]  = 1
  ... 
  >>> for key,value in mydict.items():
  ...     print(key + " " + str(value))
  ... 
  When 1
  I 2
  find 1
  myself 1
  in 4
  times 1
  of 11
  trouble 1
  Mother 2
  Mary 2
  comes 2
  to 3
  me 4
  Speaking 3
  words 7
  wisdom 7
  let 30
  it 36
  be 41
  And 3
  my 1
  hour 1
  darkness 1
  she 1
  is 4
  standing 1
  right 1
  front 1
  Let 6
  Whisper 4
  when 2
  the 4
  broken 1
  hearted 1
  people 1
  living 1
  world 1
  agree 1
  There 4
  will 5
  an 4
  answer 4
  For 1
  though 1
  they 2
  may 1
  parted 1
  there 2
  still 2
  a 2
  chance 1
  that 2
  see 1
  night 1
  cloudy 1
  light 1
  shines 1
  on 1
  Shine 1
  until 1
  tomorrow 1
  wake 1
  up 1
  sound 1
  music 1
  yeah 2
  # another way
  >>> words = s.split(" ")
  >>> for word in words:
  ...     mydict[word] = words.count(word)
  ...    
  >>> for key,value in mydict.items():
  ...    print(key, value)
  # another way
  >>> words = s.split(" ")
  >>> for word in words:
  ...     mydict[word] = 1 + mydict.get(word, 0) #利用字典自带的get方法
  ...    
  >>> for key,value in mydict.items():
  ...     print(key, value)
  ...
  

Bioinformatics Stronghold

• Counting DNA Nucleotides

  Problem

  A string is simply an ordered collection of symbols selected from some alphabet and formed into a word; the length of a string is the number of symbols that it contains.

  An example of a length 21 DNA string (whose alphabet contains the symbols 'A', 'C', 'G', and 'T') is "ATGCTTCAGAAAGGTCTTACG."

  Given: A DNA string s of length at most 1000 nt.

  Return: Four integers (separated by spaces) counting the respective number of times that the symbols 'A', 'C', 'G', and 'T' occur in s.

  # 输入一个DNA序列的字符串，统计这个DNA序列中A,C,G,T四个字符的数目，并以空格分隔输出
  >>> DNA = "TGAGTTCCGTATCGTGAAAAAGGAAGGTTAGCCAGCCTTCTCACACGATGGCTATGTTGATTTGGTCGTAGGGACACTAGATACGCTGTTTCTCTTCTTAAAGATGGAACGATGTAACGCTGGATGGAAACTGAACTAGATAAGCCAGTAGTGTAAATGTTCGTACCTGACATATCACTGTTGCAAAATTGGGGATCTATCGAAGATCAGCGCACGCGCTTCAATTGTGTTTATGGACAAGGTCTTGCGGTCGATAGGTCTTCAGACAGGCAGCGTTCCCGCGACCTTCCTCGCGATTCCTCGGATGTGGAGTGGTAGAGAACAGCGATTGGTGCCGACCTTCTGAGGAAGAACCGTGCCAGTTTGCGACATTCGTCCCAACAAATTAACCATATCACGTTAAGGTTAGCCCCCTACTAAGTACAAAGACTGGGCATCAGAAGGGCTGCCCATGTAGATGATTTGTTTTGTCAAACAAACGCGTCCGGTGGGGGCGACTTAACCACCTTCTGGCGGTGCACATTACTGGGTTCTTGAGGTGCTAATGAGCATTGCGTTGAGCGTATCAATAACGCGACTGGCGATACACCGACAGGGTATTCCGTCTGGCGCCCTGAGACGGGGTTGGGACTCCGATAAGACGTGGTGCGACCCACCGACGTGTAAAACGCAGAAGTCTCTCCTCTACATCATTGTTAAACCGGGCTAGACAGCTGCGGTCTTCTCTCATAGGCACCACCCCAGTACCCTGCCAATAGTGCATCGTGTTCTGTGAGCAAGTGTTGTTGCGGTCGCACTTGTTCTTTGTCATACAGAGGCTCCTCCTCGACTTAAGACATGTTGAAACTGTCTCGCACTGACACGCAAGGTCGGTTCACAGTTCGACCGAGCTTCAGAATGCCCAATAAAGCCTTTTCAAATAATACAACCCTGAATGCAGTCCACACCACTTTA"
  >>> mydict = {}
  >>> for item in DNA:
  ...     mydict[item] = 1 + mydict.get(item,0)
  ... 
  >>> for i in "ACGT":
  ...     print(mydict[i], end = " ")
  ... 
  237 234 240 243
  # another way
  >>> count_A, count_C, count_G, count_T = [DNA.count(nuc) for nuc in "ACGT"] #列表推导式，利用字符串自带的count方法进行计数
  >>> print(count_A,count_C,count_G,count_T)
  237 234 240 243
  # another way
  >>> count_A = DNA.count("A")
  >>> count_C = DNA.count("C")
  >>> count_G = DNA.count("G")
  >>> count_T = DNA.count("T")
  >>> print(count_A,count_C,count_G,count_T)
  237 234 240 243
  

• Transcribing DNA into RNA

  Problem

  An RNA string is a string formed from the alphabet containing 'A', 'C', 'G', and 'U'.

  Given a DNA string t corresponding to a coding strand, its transcribed RNA string uu is formed by replacing all occurrences of 'T' in t with 'U' in u.

  Given: A DNA string t having length at most 1000 nt.

  Return: The transcribed RNA string of t.

  # 输入一个DNA序列的字符串,将其转换为RNA序列进行输出
  >>> DNA = "GTTTTGTCTAAGGCCCCGAAAATGATCTTGGACACCCCTCTCCCGATTTTCACTGCGATTTCACTTTTCATCCGGACGCCTGGTAGCGCAGGTATAAACTAAGGAGCAAGACGCTTGAAAGTTGTTAGGATAGATGCGGCCATGCCATATGCTGATATCCAGATGGTCGGTTTAACTTTATTTGGTCGAACCGTAGTACGGGGGAAACCGACCATAATGTCTATTGATTCCAGGTGTTAACCCATCAAACGCAGGCCAACATTCTCCGTATTGAGTTCACCCATCGATCTCACAGCGGTTAGGAACACGAGTGATCTAGTCGGCACTTACGCGCCGCTAGCGGGGTGCTCTACGAGTATTGCATAATGACGAGATAAAATGGGACCGTATCAATCCGGATTTCTAACGACTTGCTCCCAAAGGATACGTTACGTAATTAACTCAGGCATAGGCGTCATGGCATGTAATTTACTATGTCGGGCGCACGTAGTAATCCCTACCAAGCAGCGTAGACTAGCGAAAGTCCCCTGGTACAGTATTGTTCTTCTGAAGTCGCGTACTCCCCCAGTTCCTAGAAAAATGGGTAAATGGACTCTCGCTCGTCTCCTCGCTGGTTGGTCATGGCGTTGGGTCTGCTTTACCCCTCCAAGCCGCACGTAGCCGCCTACAGTAAATCGGGGGTGCTGGCTTACCGACGTTAAGTTGCTCCAACTACGGTGGACTGGTATCACCTACTAGTTCCCTATAGCGCGTTCTGGCTGGTTCCTAGCTGGGCACTAGGTACGGGAATTCCCCCGTACCAGACTAAAAGGAATGTAGAACGTGATGGGCATTGACACTGAACTCCTTTTTCGGGCCGGTCAGGTTGCATGGGCAACGGCCTACAACGACGTCAAACTCACCATGGCGGGGTTCCTGGGTATGCGTGCCGCTCAGGTTCTACCCTCTACCAACGAGGACGCCAACAACTGATTAC"
  >>> RNA = []
  >>> for item in DNA:
  ...     if item == "T":
  ...             item = "U"
  ...             RNA.append(item)
  ...     else:
  ...             RNA.append(item)
  ... 
  >>> print("".join(RNA))
  GUUUUGUCUAAGGCCCCGAAAAUGAUCUUGGACACCCCUCUCCCGAUUUUCACUGCGAUUUCACUUUUCAUCCGGACGCCUGGUAGCGCAGGUAUAAACUAAGGAGCAAGACGCUUGAAAGUUGUUAGGAUAGAUGCGGCCAUGCCAUAUGCUGAUAUCCAGAUGGUCGGUUUAACUUUAUUUGGUCGAACCGUAGUACGGGGGAAACCGACCAUAAUGUCUAUUGAUUCCAGGUGUUAACCCAUCAAACGCAGGCCAACAUUCUCCGUAUUGAGUUCACCCAUCGAUCUCACAGCGGUUAGGAACACGAGUGAUCUAGUCGGCACUUACGCGCCGCUAGCGGGGUGCUCUACGAGUAUUGCAUAAUGACGAGAUAAAAUGGGACCGUAUCAAUCCGGAUUUCUAACGACUUGCUCCCAAAGGAUACGUUACGUAAUUAACUCAGGCAUAGGCGUCAUGGCAUGUAAUUUACUAUGUCGGGCGCACGUAGUAAUCCCUACCAAGCAGCGUAGACUAGCGAAAGUCCCCUGGUACAGUAUUGUUCUUCUGAAGUCGCGUACUCCCCCAGUUCCUAGAAAAAUGGGUAAAUGGACUCUCGCUCGUCUCCUCGCUGGUUGGUCAUGGCGUUGGGUCUGCUUUACCCCUCCAAGCCGCACGUAGCCGCCUACAGUAAAUCGGGGGUGCUGGCUUACCGACGUUAAGUUGCUCCAACUACGGUGGACUGGUAUCACCUACUAGUUCCCUAUAGCGCGUUCUGGCUGGUUCCUAGCUGGGCACUAGGUACGGGAAUUCCCCCGUACCAGACUAAAAGGAAUGUAGAACGUGAUGGGCAUUGACACUGAACUCCUUUUUCGGGCCGGUCAGGUUGCAUGGGCAACGGCCUACAACGACGUCAAACUCACCAUGGCGGGGUUCCUGGGUAUGCGUGCCGCUCAGGUUCUACCCUCUACCAACGAGGACGCCAACAACUGAUUAC
  # another way
  >>> RNA = DNA.replace("T","U") #利用字符串自带的replace方法进行替换
  >>> print(RNA)
  

• Complementing a Strand of DNA

  Problem

  In DNA strings, symbols 'A' and 'T' are complements of each other, as are 'C' and 'G'.

  The reverse complement of a DNA string ss is the string scsc formed by reversing the symbols of ss, then taking the complement of each symbol (e.g., the reverse complement of "GTCA" is "TGAC").

  Given: A DNA string s of length at most 1000 bp.

  Return: The reverse complement sc of s.

  # 输入一个DNA序列的字符串,将其转换为反向互补序列进行输出
  >>> DNA = "TGCAGCACACTGGGTCGGGCTTTTGCTTGTATTATCTCACATAAGGGTAGCATAAACCTCTTAAATCGGTCAGTTTTCCTTGGAGTAGGTTGTTACCATGGGCAAACCATTCCCAGTGGTAAGATACCGAGCAGCGTACGCGACAGGCTGTCTGATCGGACAATCGGAGTACAGAACCACCCATTTGAAACACATGCTTTAGCGCTTGGCCCTCACGAAGCCCCATACAGTCACCACCCTCTAATCTTTGGCGTTACTGTGCAGGCGCCGAGAGATTAAACTTTCAATAGTGAGGGCCTAAAACCCTATCTTAGTGACTAACCAGATATATCGTTCATTCTCATTCCGTGCCCCCCAACCTCATTATAATTACCCGAAGAGTTAACGCCTTGAGAACATCTTCACGCGGGAGCACAAGTCCGAGACTGACACCGACGCAAATGGGTGATTCTTTGGGTCACGGATCGGGATTGCCGGAAACCAATAAATTAATTACTTATACCGTCCTCGTATTCGTTCCCCCAATGCGAGCTCAAAACTATAACTAATGAAGTTTTCCATTAACCCGGTCCATACATCATCGTTTCAGGACAAAAAATTTCAGGGAGCAGCCTCACTTGTGCTAGCCTAAAAATTACATCTTTGATACACATAGTAATTGGTGATATTTAGCTGCCTCCAGCACCGGTGGAGCACAGTCATGCCGAGTGTTAGACGCCATAAGGGAGTCGGCAATATAGACGAGCCATTTTTACAAATATGCGTTGGTCGACTACCATTGCTCGGAGGACAGGCTGTTTTTACCAAAGGGCGCTCTCATGAATCCCCACACCACTAGCGAACTTGTCGAACTTC"
  >>> mydict = {"A":"T","T":"A","C":"G","G":"C"}
  >>> DNA_revc = []
  >>> for item in DNA:
  ...     if item in mydict:
  ...             DNA_revc.append(mydict[item])
  ... 
  >>> DNA_revc.reverse() # 利用列表自带的reverse函数对原字符串进行反转
  >>> print("".join(DNA_revc))
  GAAGTTCGACAAGTTCGCTAGTGGTGTGGGGATTCATGAGAGCGCCCTTTGGTAAAAACAGCCTGTCCTCCGAGCAATGGTAGTCGACCAACGCATATTTGTAAAAATGGCTCGTCTATATTGCCGACTCCCTTATGGCGTCTAACACTCGGCATGACTGTGCTCCACCGGTGCTGGAGGCAGCTAAATATCACCAATTACTATGTGTATCAAAGATGTAATTTTTAGGCTAGCACAAGTGAGGCTGCTCCCTGAAATTTTTTGTCCTGAAACGATGATGTATGGACCGGGTTAATGGAAAACTTCATTAGTTATAGTTTTGAGCTCGCATTGGGGGAACGAATACGAGGACGGTATAAGTAATTAATTTATTGGTTTCCGGCAATCCCGATCCGTGACCCAAAGAATCACCCATTTGCGTCGGTGTCAGTCTCGGACTTGTGCTCCCGCGTGAAGATGTTCTCAAGGCGTTAACTCTTCGGGTAATTATAATGAGGTTGGGGGGCACGGAATGAGAATGAACGATATATCTGGTTAGTCACTAAGATAGGGTTTTAGGCCCTCACTATTGAAAGTTTAATCTCTCGGCGCCTGCACAGTAACGCCAAAGATTAGAGGGTGGTGACTGTATGGGGCTTCGTGAGGGCCAAGCGCTAAAGCATGTGTTTCAAATGGGTGGTTCTGTACTCCGATTGTCCGATCAGACAGCCTGTCGCGTACGCTGCTCGGTATCTTACCACTGGGAATGGTTTGCCCATGGTAACAACCTACTCCAAGGAAAACTGACCGATTTAAGAGGTTTATGCTACCCTTATGTGAGATAATACAAGCAAAAGCCCGACCCAGTGTGCTGCA
  # another way
  >>> mydict = {"A":"T","T":"A","C":"G","G":"C"}
  >>> DNA_revc = []
  >>> for item in reversed(DNA): # 利用reversed函数对字符串进行反转，或着反向索引DNA[::-1]进行反转
  ...     DNA_revc.append(mydict[item])
  ...             
  >>> print("".join(DNA_revc))